Answer
The radius of convergence for the power series is $R=\infty$ and the interval of convergence is $(-\infty, \infty)$.
Work Step by Step
Root Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty} \sqrt[k] a_k$
1) When $0 \leq l \lt 1$. then the series $\Sigma a_k$ converges
2) When $l \gt 0$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\sin^k (\dfrac{1}{k}) x^k$
Now, $l=\lim\limits_{k \to \infty} \sqrt[k] {\sin^k (\dfrac{1}{k}) x^k}=|x| \lim\limits_{k \to \infty} |\sin \dfrac{1}{k}|=0$
So, the radius of convergence for the power series is $R=\dfrac{1}{l}=\dfrac{1}{0}=\infty$. This implies that the interval of convergence is $(-\infty, \infty)$.
